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3 years ago

The number of text messages (t) that katie sends depends on the number of days (d) she is on vacation. the equation is t=50d+20.

1 answer:

3 0

Based on the equation katie gets 20 text messages + 50 more for each day she is away.
therefore the number 50 shows how much the number of text messages (t) increases for each day she is away.

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Tiffany used to live 15 kilometers away from the school, but after she moved she now lives 9.9 kilometers away. She is _____ per

valkas [14]

Answer:

Tiffany is 34% closer to the school.

Step-by-step explanation:

Given the statement: Tiffany used to live 15 kilometers away from the school, but after she moved she now lives 9.9 kilometers away.

She lived away from the school = 15 km

as, she moved 9.9 km away.

Now, the distance closer to school from where she lived = 15 -9.9 = 5.1 km

To, find how much percent she is closer to school.

Percent states that a number or ratio expressed as a fraction of 100.

$percent = \frac{5.1}{15} \times 100$ = $\frac{51 \times 100}{150} = \frac{51 \times 10}{15}$

Simplify:

percent closer to school = 34%

Therefore, she is 34% closer to the school.

5 0

4 years ago

zepelin [54]

The question is incomplete. Below you will find the missing figure.

Two friends, Andrew and Liz, are playing a game using this spinner. If the

spinner lands on region 2 or 3, Andrew wins. If it lands on region 4, Liz wins. If it lands on any other region, neither wins. Is this a fair game?

A. Yes. Each friend has a 1/4 probability of winning.

B. Yes. Each friend has a 1/6 probability of winning.

C. No. Andrew has a probability of 1/4 winning and Liz has a probability of 1/6 winning.

D. No. Andrew has a probability of 1/3 winning and Liz has a probability of 1/4 winning.

The correct option is Option A: Yes. Each friend has 1/4 probability of winning.

Probability is the possibility of an event to happen which is the ratio of no. favorable outcomes and no. of total outcomes.

P = no. of favorable outcome/ total no. of outcome

Here given in the question,

If the spinner touches region 2 or 3, Andrew wins

If the spinner touches region 4, Liz wins.

This is fair play because both have the same size of region for winning.

If we cut the circle into 8 pieces,

as region 1 is the sum of 2 sub-parts of the circle,

possibility of spinner touching region 1= 2/8= 1/4

possibility of spinner touching the small region = 1/8

So the possibility of the spinner touching the region 2or 3= 1/8+1/8= 2/8= 1/4

Hence the possibility of Andrew wins= possibility of the spinner touching region 1= 1/4

possibility of Liz wins= possibility of the spinner touching the region 2 or 3= 1/4

Therefore the correct option is Option A: Yes. Each friend has 1/4 probability of winning.

Learn more about the probability

here: brainly.com/question/24756209

#SPJ10

6 0

2 years ago

Fernando had 2 children. Each of their children had 2 children, and each of these

rjkz [21]

Answer:

Step-by-step explanation:

Fernando-

Child 1- Child 2-

2 kids- 2 kids- = 4 grandchildren for Fernando

2 kids 2 kids These are his great grandchildren, not grandchildren.

7 0

3 years ago

Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving

lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

$P(x=0) = ^nC_x P^x (1-P)^n^-^x$

$P(x=0) = ^5C_0 (0.4)^0 (1-0.4)^5^-^0$

$P(x=0) = ^5C_0 (0.4)^0 (0.60)^5$

$P(x=0) = 0.778$

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

$= 1 - P(X = 0)$

$= 1 - ^5C_0 (0.4)^0 * (0.6)^5$

$= 1 - 0.0778$

$= 0.9222$

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

$^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 (0.4)^2 (0.6)^3$

$= 0.6826$

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

$= 1 - [^5C_0 (0.4)^0 (0.6)^5 + ^5C_1 (0.4)^1 (0.6)^4 + ^5C_2 * (0.4)^2 (0.6)^3]$

$= 1 - 0.6826$

$= 0.3174$

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0

3 years ago

What is not an acceptable form of proof

STALIN [3.7K]

No reason, or no evidence to support your claim/proof

4 0

3 years ago